题目：

题解：

void swap(int * nums,int indexA,int indexB)
{int temp    = nums[indexA];nums[indexA]= nums[indexB];nums[indexB]= temp;
}void prem(int* nums, int numsSize, int* returnSize, int** returnColumnSizes,int** returnNums,int offset)
{if(offset == numsSize){//遍历到末尾了//申请returnNumsreturnNums[*returnSize] = (int *)malloc(sizeof(int ) * numsSize);//拷贝内容到returnNumsmemcpy(returnNums[*returnSize],nums,sizeof(int) * numsSize );//记录当前拷贝内容的长度(*returnColumnSizes)[*returnSize] = numsSize;*returnSize = *returnSize + 1;}else{//回溯算法的核心int i;for(i = offset; i < numsSize; i++){swap(nums,i,offset);//i 和 offset 交换prem(nums,numsSize,returnSize,returnColumnSizes,returnNums,offset+1);swap(nums,i,offset);//i 和 offset 交换}}
}int** permute(int* nums, int numsSize, int* returnSize, int** returnColumnSizes)
{//不重复的数字的全排序//组合次数为 n！= n *( n - 1) *( n - 2) ...... 2 * 1//这样的方法适合回溯的方法//取值范围1 <= nums.length <= 6  = 6 * 5 * 4 * 3 *2 * 1 = 720中可能int **returnNums = (int **)malloc(sizeof(int *) * 721);*returnColumnSizes= (int *)malloc(sizeof(int ) * 721);*returnSize = 0;prem(nums,numsSize,returnSize,returnColumnSizes,returnNums,0);return returnNums;}